Solve (sqrt{8}+sqrt{15}+sqrt{8}-sqrt{15})^2

$$( \sqrt { 8 } + \sqrt { 15 } + \sqrt { 8 } - \sqrt { 15 } ) ^ { 2 }$$

Evaluate

$$32$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\left(2\sqrt{2}+\sqrt{15}+\sqrt{8}-\sqrt{15}\right)^{2}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\left(2\sqrt{2}+\sqrt{15}+2\sqrt{2}-\sqrt{15}\right)^{2}$$

Combine 2\sqrt{2} and 2\sqrt{2} to get 4\sqrt{2}.

$$\left(4\sqrt{2}+\sqrt{15}-\sqrt{15}\right)^{2}$$

Combine \sqrt{15} and -\sqrt{15} to get 0.

$$\left(4\sqrt{2}\right)^{2}$$

Expand \left(4\sqrt{2}\right)^{2}.

$$4^{2}\left(\sqrt{2}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$16\times 2$$

Multiply 16 and 2 to get 32.

$$32$$

Factor

$$2^{5}$$